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Unformatted text preview: STT 315 (Sec 1 — 11) Exam #1  FORM A Sb \Hn 003 Mon, __ a . Gilliland (Closed book  75 minutes) :30 am exam Name PID 0. You may use a calculator. All electronic devices capable of communication and/or web access must be turned off.
1. In the places above, please put your name and PID. Tables are found on the last sheet. 2. Be sure to code your name, section number and PID number correctly on the answer sheet that has been provided.
Sign the answer sheet. 3. On your answer sheet you must also code the letter A as the FORM of your examination in the space provided. 4. After the exam, return your answer sheet as you leave. You may keep your exam paper.
5. There are 30 questions, each worth 1.667 course points. (Total 50. 010 course points possible.) 6. Your score is determined from your answer sheet and not from what you have written on your exam paper. 1  2. A deck of cards consists of 52 cards (4 are Jacks, 4 are Queens, 4 are Kings and 40 are other cards). We refer to the
Jacks, Queens and Kings as the face cards. Three cards are selected at random from the deck one after the other without
replacement. ' 1. What is the probability of the cute in 2 My! ic‘t‘i‘ﬂm
Ra i €
2. What is the probability of getting at east one face card in the three selections? (a) .36 (b) .40 (c) .64 (e) .70 t..uo as as Jack, Queen, King) in that order? 4 3 2 (b) 0523 (c) .001 (d) ———— (e) lisici 21;; 1,000 claims to Medicare are cross—classiﬁed by the location of the patient (East or West) and by the Type of Claim
(Hospitalization, Physician’s Visit, Outpatient Treatment). Below is a table with the counts of the claims in the various
categories. From the 1,000 claims, a claim is selected at random. For example, the event East refers to outcomes where the
claim is for a patient item the East; thus, P(East) m 600/1000 = .60. East West Total
Hospitalization 600
Physician's Visit 300
Out I atient Treatment 100
Total 600 400 1000
3. P(East U Hospitalization) = (b) 1.20 (c) .60 (d) .20 (e) .40
e800 _/0 \
4. P030151?  Hospitalization) = (b) .25 (c) .60 (d) .55 (e) .10
5a;
660
5. P(West u Hospitalization): (a) .50 (b) .65 (d) .35 (e) .70 PCWUH}: .30 ea WW5: .233 _ 6 — 7. You start interviews for a job, one interview after the other, until you get an offer for the ﬁrst time. Use the model
where results of interviews (Offer, No Offer) are independent with chance of Offer equal to .30 on each interview to answer these questions. . (do n N
6. What is the probability that you get need more than 4 interviews to get your ﬁrst Job offer? “if: “fin a; (a) .33 (c) .76] (d) .54 (e) .50 7. What is the expected number of interviews to get your ﬁrst offer? (a)3.73 (b)3.65 (c) 3.80 (d) 4.12 (Ee; 3.33 s__9. For a particular probability model, P(A) = .6, P(B) = .3, P(A  B) = .5. i‘ C An 33 '' 9335300103)
8. P(A u B) = r __ (b) .55 (c) .60 7 (d) .90 (e) .30
. a + " o l a
9. Are events A and B independent? (a) Yes 3 l in be 5"": é l $9 % 10  11. A class consists of 60% men and 40% women. Of the men, 25% are blond, while 50% of the women are blond. A
student is chosen at random from the class. 10. What is theprobability that the chosen student is blond? (a) .50 (b) .75 (d) .25 (e) .70 Hm séeﬂtmm ens7345,50}: .5§ 11. If the chosen student is found to be blond, what is the probabilityat student is a man? (a) .57 (c) .50 (d) .67 (e) .23 P004013.) .3 Mama) ,3... “3‘3; _
@(Mligﬁw «W ~ W ‘33. Milli era) ‘3‘3 12 — 14. Here is the probability mass function of a discrete random variable X. “1
I.
12. What is the value of the cumulative distribution ﬁlnction at x x 3?
(a) .20 (b) .70 (c) .80 (e) 1.00 F63) :Pl’xea) = 8' 0 13. What is the expected value ofX? (b) 2.00 (c) 1.60 (d) 1.85 (e) 1.00 omen 1636} + 2 (pi+362»; +4616) :1 1.30 (b) .60 (c) .40 (d) .30 (e) .70 905 + pal1P“) "3 .673 _l_5. A random variable X has a mean of 50 and a standard devia '  a e . Then P(30 < X < 70) is at least
(a) 3/9 @ (c) .95 (d) 15/16 Cheloyslted W'Hl 5‘21 14. P(15X<4)=
(6) 24/25 16 — 17. A large pOpulation of parts is 10% defective. A random sample of 20 parts is to be drawn from the population. Let
X denote the number of defective parts in the sample. (a) .500 @ (e) .609 (d) .677 (e) .285 E); ﬁblel an "FT32: PKXtLX—P(Xéo)=,g77,ua
’ "5.663" 16. P(0<X52)= XNBQOJW) 17. Determine 1301) and 31301). (a)3.5, 1.56 (b)2,1.00 (d) 4, 1.34 (0) 2,2 Em, 206101: 2. 85300:“ 1001016301 =01»? :IM 18 — 19. A small population consists of 3 defective parts and 27 good parts (30 parts in total). A simple random sample of
20 arts is to be drawn from the population. Let X denote the number of defective parts in the sample. (a) .54 (b) .28 (c) .66 (d) .15 (e) .05
3) Z 7
3 . 7 :: 18. P{X=3)= ( a l x 644362.36“ I‘ ’ ’ #:3271903
ﬂ 30,005,015’“
. 20 19. Determine E(X) and SD(X). Em): 306103 :2. 5m.) 2:" \i (a) 3.5, 1.46 (c) 2,1.34 z. \i @3915 3’ Q??? (d) 2,1.46 (e) 2,2 E. The time X of arrival of a customer to an appointment is modeled as a continuous random variable with a uniform
distribution on the interval from 10 minutes late to 30 minutes late, say from 10 to 30. What is the probability that X is less than 15?
(a) .40 (b) .15 (c) .60 (d) .75
_L , ‘ ,, \SIO m5; :'
Lobm—a y P(XHSX " 3040 *9 2g
0 to 20 30 a. The time to failure X for a part under accelerated testing is modeled as continuous Exponential random variable with expected value (mean) 4 hours. Here is a plot of the cumulative distribution function y = F(x) = 1 ~— e'xm. Il==l=:!!!!===ii
Ilial IIIIII
=aa..= W_Hlllll
synIII What is the probability of failure within the ﬁrst two hours? ' I
. (a) .50 (b) .78 (d) .35 (e) .65 P(Xez‘g : “(2‘) =—*" r40 22  23. A variable is modeled as a random variable X with a normal distribution with mean 40 and standard deviation 5. (b) .95 (c) .68 Uae at“ (e) 45.7 wag, Ti sirmick? 2.. 22. P(35 < X < 50) = (d) .997 (e) .74 23. Find the 75th percentile of the distribution, that is, the point x such that P(X S x) = .75.
(a) 44.7 (b) 42.8 (c) 46.2 a. A distribution of raw scores x has a normal distribution with mean 100 and standard deviation 10. What is the standard score (zscore) of the raw score x i 116?
(a) 1.6 (b) 11.6 (c) .6 (d) 1.0 (e) .70 “64”; 1.5
[0 2_5. Calculate the chance of at least 1,750 successes in n = 3,600 independent trials where the chance of success is .50 on each trial? .
(a) .997 (b) .85 (c) .68 (e) .78 X1" M04)? Successes I
P(><> V750) a” 1 “We 1747‘) r“ 1 binomcﬁi‘c’ssag.5,i7ta) l” a 046 :3": “1591‘
Or use manual 71(3ng _
i7HQ.5—tsae L. _
Em: Essen pm; 174%“) :PC as. so ) We; 1,7,9
5D“): 50 ﬁ.'5+ii‘i.555 5.4552? E. In applications, selecting a simple random sample from a population may not be a simple task. . .—
II (b) False _2_j_. The formulas that we are using for the standard deviation of)? are based on the sample drawn with simple random
sampling. " '  (b) False E. The genders and ages of the 5 children makingup a population are {F3, M3, M4, F6, M10}. For example, F3 refers to a female age 3. A simple random sample of size 2 is to be drawn from the 5 children. Let f) be the sample proportion of
Males M. (e) .70 S 32. Suppose that x in units of feet is normally distributed across a large population of interest with unknown mean ,u and
known standard deviation cr= 8 feet. Suppose that a random sample of size n = 16 is to be selected from a population. Use a
normal distribution for a? to determine the probability that the sample average 3? will fall within 2 feet of [J to two decimals. (a) .95 (b) .38 (c) .87 (e) none of these E. A large population of trees is 20% diseased. The population is sampled with a simple random sample of size 100. What
is the probability that the sample of trees has at least 15 diseased trees in it, that is, that the sample proportion is at least .15?
If you wish, you may use normal approximation to determine P( p“ Z .15). Pick the closst. .9... (a) .76 (b) .65 (d) .30 (e) .50 ...
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This note was uploaded on 02/01/2011 for the course STAT 315 taught by Professor Dennisgilliland during the Summer '10 term at Michigan State University.
 Summer '10
 DennisGilliland

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